If ( x + iy ) ( 3 – 4i ) = (5 + 12i ) , then root of(x2 + y2) =??

Dear Student,

$$\begin{gathered} (\mathrm{x}+\mathrm{iy})(3-4\mathrm{i}) = (5 +12\mathrm{i} ) \\ \Rightarrow 3\mathrm{x}-4\mathrm{ix}+3\mathrm{iy}+4\mathrm{y}=5+12\mathrm{i} \\ \Rightarrow (3\mathrm{x}+4\mathrm{y})+\mathrm{i}(-4\mathrm{x}+3\mathrm{y})=5+12\mathrm{i} \end{gathered}$$
Now equating real and imaginary parts , we get

$$\begin{gathered} 3\mathrm{x}+4\mathrm{y}=5 .....\left(1\right) \mathrm{and} -4\mathrm{x}+3\mathrm{y}=12 .....\left(2\right) \\ \mathrm{Multiply} \left(1\right) \mathrm{by} 4 \mathrm{and} \left(2\right) \mathrm{by} 3, \mathrm{we} \mathrm{get} \\ 12\mathrm{x} + 16\mathrm{y} = 20 ....\left(3\right) \\ -12\mathrm{x} + 9\mathrm{y} = 36 ....\left(4\right) \\ \mathrm{Adding} \left(3\right) \mathrm{and} \left(4\right), \mathrm{we} \mathrm{get} \\ 25\mathrm{y} = 56 \\ \Rightarrow \mathrm{y} = \frac{56}{25} \\ \mathrm{Now}, \mathrm{from} \left(1\right), \mathrm{we} \mathrm{get} \\ 3\mathrm{x} + \frac{224}{25} = 5 \\ \Rightarrow 3\mathrm{x} = 5 - \frac{224}{25} \\ \Rightarrow 3\mathrm{x} = -\frac{99}{25} \\ \Rightarrow \mathrm{x} = -\frac{33}{25} \end{gathered}$$

So we get,

$$\begin{gathered} \mathrm{x}=\frac{-33}{25} \mathrm{and} \mathrm{y}=\frac{56}{25} \\ \mathrm{Now}, ({\mathrm{x}}^2+{\mathrm{y}}^2)={\left(-\frac{33}{25}\right)}^2+{\left(\frac{56}{25}\right)}^2=\frac{4225}{625} \\ \Rightarrow \sqrt{({\mathrm{x}}^2+{\mathrm{y}}^2)}=\sqrt{\frac{4225}{625}}= \frac{65}{25} = \frac{13}{5} = 2.6 \end{gathered}$$

Regards.